Question

# The line 2x + 3y = 12 meets the x-axis at A and y-axis at B. The line through (5, 5) perpendicular to AB meets the x-axis and the line AB at C and E respectively. If O is the origin of coordinates, find the area of figure OCEB.

Solution

## The given line is 2x + 3y = 12, which can be written as $\frac{x}{6}+\frac{y}{4}=1$              ... (1) So, the coordinates of the points A and B are (6, 0) and (0, 4), respectively. The equation of the line perpendicular to line (1) is $\frac{x}{4}-\frac{y}{6}+\lambda =0$ This line passes through the point (5, 5). $\therefore \frac{5}{4}-\frac{5}{6}+\lambda =0\phantom{\rule{0ex}{0ex}}⇒\lambda =-\frac{5}{12}$ Now, substituting the value of $\lambda$ in $\frac{x}{4}-\frac{y}{6}+\lambda =0$, we get:          Thus, the coordinates of intersection of line (1) with the x-axis is . To find the coordinates of E, let us write down equations (1) and (2) in the following manner: $2x+3y-12=0$        ... (3) $3x-2y-5=0$          ... (4) Solving (3) and (4) by cross multiplication, we get: Thus, the coordinates of E are (3, 2). From the figure, $EC=\sqrt{{\left(\frac{5}{3}-3\right)}^{2}+{\left(0-2\right)}^{2}}=\frac{2\sqrt{13}}{3}$ $EA=\sqrt{{\left(6-3\right)}^{2}+{\left(0-2\right)}^{2}}=\sqrt{13}$ Now, MathematicsRD Sharma XI (2015)Standard XI

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